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Suppose $X_n$ is a sequence of random variables with $|X_n| \leq 1$. Assume $X_n \overset{p}{\to} X$. True or false: $Var(X_n)\to 0$. If false, give a counterexample. If true, give an argument.

Convergence of $\operatorname{Var}(X_n)$ if $|X_n| \le 1$ seems relevant. How much does the $X_n \to 0$ change what I have to say? Would appreciate a start in the right direction thanks!

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Suppose $X_n$ is a sequence of random variables with $|X_n| \leq 1$. Assume $X_n \overset{p}{\to} X$. True or false: $Var(X_n)\to 0$. If false, give a counterexample. If true, give an argument. \

\indent The statement is false. Here we can offer a counter example.\ \ \indent Let $X \sim U[0,1]$. Since $X_n \overset{p}{\to} X$ we know $X_n \overset{d}{\to} X$. This implies that $Var(X_n) = Var(X)$. Using the definition of the continuous uniform distribution we can say that $f_X(x) = \frac{1}{1-0} = \frac{1}{1}$. We can then follow the steps outlined here and say:

\begin{equation*} \begin{aligned} & \operatorname{Var}(X) = \int_{-\infty}^\infty x^2 f_X(x) \, dx - \left( E(X) \right)^2 \\ & = \int_{-\infty}^0 0 f_X(x) \, dx + \int_0^1 x^2 f_X(x) \, dx + \int_0^\infty 0 f_X(x) \, dx - \left( \frac{0+1}{2} \right)^2 \\ & = \int_0^1 x^2 (1) \, dx - \frac{1}{4} \\ & = \left( \frac{x^3}{3} \right) \bigg\rvert_0^1 \\ & = \left( \frac{1}{3} - \frac{0}{3} \right) - \frac{1}{4} \\ & = \frac{1}{12} \end{aligned} \end{equation*}\

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    $\begingroup$ $Var(X_n) \to Var(X)$ is a more sensible conclusion than the one you propose. In fact, there's an obvious counterexample to the statement you make, take all the $X_i$ equal to some non-constant distribution, like $U[0,1]$. Then $X$ is also $U[0,1]$ but $Var(X_n)$ is the same non-zero number. $\endgroup$ Commented Oct 19, 2020 at 3:49
  • $\begingroup$ I'm going to try to edit this in a second with an answer in this form. Could you look back and let me know what you think? $\endgroup$ Commented Oct 19, 2020 at 4:15
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    $\begingroup$ Definitely, I'd like to be of help to you : and your answer is fine. $\endgroup$ Commented Oct 19, 2020 at 4:27
  • $\begingroup$ It is, however, a little long-winded if you like. The point is, that if a random variable is non-constant (almost surely) then its variance is non-zero. So you could have just used this fact instead of actually computing the variance of $U[0,1]$ (which is also correctly done, good job) and saying it is non-zero. Also, you should write $X_n \sim U[0,1]$ for all $n$ , because you have used it. This works as a counterexample. Finally, it is possible to prove for this question that $Var(X_n) \to Var(X)$, in the same way as the post you attached. $\endgroup$ Commented Oct 19, 2020 at 4:29
  • $\begingroup$ Awesome thanks! I really appreciate your help. Its hard working during pandemic and not being able to ask peers for help. Means a lot for you to take the time. Thanks $\endgroup$ Commented Oct 19, 2020 at 4:32

2 Answers 2

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For your first question - obviously false. Let $X_m=X_n$ for all indices, so they converge trivially to the same random variable with the same non-zero variance.

I suspect that for the second question, the variance would converge to the variance of the limit, but I haven't tried to work it out.

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    $\begingroup$ The variance would converge to the variance of the limit (+1), in fact convergence is distribution along with a.s. boundedness is enough for this. It follows from the Portmanteau theorem, and the fact that the $X_i$ are bounded so all moments exist. $\endgroup$ Commented Oct 19, 2020 at 4:41
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I added an edit that posted my answer after following some advice given! Thanks to Teresa!

Let $X \sim U[0,1]$. Since $X_n \overset{p}{\to} X$ we know $X_n \overset{d}{\to} X$. This implies that $X_n \sim U[0,1]$ $Var(X_n) = Var(X)$ and we will show this $\neq 0$ for at least the following case. Using the definition of the continuous uniform distribution we can say that $f_X(x) = \frac{1}{1-0} = \frac{1}{1}$. We can say:

\begin{equation*} \begin{aligned} & Var(X) = \int_{-\infty}^{\infty} x^2 f_X(x) dx - \left( E(X) \right)^2 \\ & = \int_{-\infty}^{0} 0 f_X(x) dx + \int_{0}^{1} x^2 f_X(x) dx + \int_{0}^{\infty} 0 f_X(x) dx - \left( \frac{0+1}{2} \right)^2 \\ & = \int_{0}^{1} x^2 (1) dx - \frac{1}{4} \\ & = \left( \frac{x^3}{3} \right) \bigg\rvert_{0}^{1} \\ & = \left( \frac{1}{3} - \frac{0}{3} \right) - \frac{1}{4} \\ & = \frac{1}{12} \end{aligned} \end{equation*}\

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    $\begingroup$ One last thing : that attempt part of your question can be posted here as well : that is the answer, so it should come here. +1 anyway. $\endgroup$ Commented Oct 19, 2020 at 4:40
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    $\begingroup$ Just seen the edit, thanks! $\endgroup$ Commented Oct 19, 2020 at 4:49

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